[其它] 程序运行速度疑问 [复制链接]

zhangxujun1981

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11楼 [报告]

发表于 2010-10-25 14:14 |只看该作者

本帖最后由 zhangxujun1981 于 2010-10-25 14:20 编辑

void TestSysCallSpeed(void)
{
int i;

for(i=0;i<9000000;i++){
(*(volatile short *)(GBankAddr)=0);
(*(volatile short *)(GBankAddr)=1);
(*(volatile short *)(GBankAddr)=0);
(*(volatile short *)(GBankAddr)=1);
(*(volatile short *)(GBankAddr)=0);
(*(volatile short *)(GBankAddr)=1);
}
}

Test.o:    file format elf32-littlearm

Disassembly of section .text:

00000f40 <TestSysCallSpeed>:
   f40: e59f3040 ldr r3, [pc, #64] ; f88 <TestSysCallSpeed+0x48>
   f44: e59f2040 ldr r2, [pc, #64] ; f8c <TestSysCallSpeed+0x4c>
   f48: e5933000 ldr r3, [r3]
   f4c: e3a01000 mov r1, #0 ; 0x0
   f50: e1c310b0 strh r1, [r3]
   f54: e3a01001 mov r1, #1 ; 0x1
   f58: e1c310b0 strh r1, [r3]
   f5c: e3a01000 mov r1, #0 ; 0x0
   f60: e1c310b0 strh r1, [r3]
   f64: e3a01001 mov r1, #1 ; 0x1
   f68: e1c310b0 strh r1, [r3]
   f6c: e3a01000 mov r1, #0 ; 0x0
   f70: e1c310b0 strh r1, [r3]
   f74: e2522001 subs r2, r2, #1 ; 0x1
   f78: e3a01001 mov r1, #1 ; 0x1
   f7c: e1c310b0 strh r1, [r3]
   f80: 1a0003d1 bne 1ecc <TestWinDr+0x16c>
   f84: e1a0f00e mov pc, lr
   f88: 00000000 andeq r0, r0, r0
   f8c: 00895440 addeq r5, r9, r0, asr #8

-------------------------------------------------------------------------------

void TBank(void)
{

(*(volatile short *)(GBankAddr)=0);
(*(volatile short *)(GBankAddr)=1);
(*(volatile short *)(GBankAddr)=0);
(*(volatile short *)(GBankAddr)=1);
(*(volatile short *)(GBankAddr)=0);
(*(volatile short *)(GBankAddr)=1);

}

t.o:    file format elf32-littlearm

Disassembly of section .text:

00000000 <Tbank>:
0: e59f1038 ldr r1, [pc, #56] ; 40 <.text+0x40>
4: e3a00001 mov r0, #1 ; 0x1
8: e3a0c000 mov ip, #0 ; 0x0
c: e5913000 ldr r3, [r1]
  10: e1c300b0 strh r0, [r3]
  14: e5912000 ldr r2, [r1]
  18: e1c2c0b0 strh ip, [r2]
  1c: e5913000 ldr r3, [r1]
  20: e1c300b0 strh r0, [r3]
  24: e5912000 ldr r2, [r1]
  28: e1c2c0b0 strh ip, [r2]
  2c: e5913000 ldr r3, [r1]
  30: e1c300b0 strh r0, [r3]
  34: e5912000 ldr r2, [r1]
  38: e1c2c0b0 strh ip, [r2]
  3c: e12fff1e bx lr
  40: 00000000 .word 0x00000000

第一个是应用程序,第二个是驱动程序.好像没有什么区别.可能是你上面说的关于cache 使用的不一样吧.我马上试一试.
像for(i=0;i<50;i++)这样运行很快是意料之中的事情,cpu的主频是667M,而且是8级流水线的arm11

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zhangxujun1981

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12楼 [报告]

发表于 2010-10-25 14:24 |只看该作者

void TestSysCallSpeed(void)
{
int i;

for(i=0;i<9000000;i++){
(*(volatile short *)(GBankAddr)=0);
(*(volatile short *)(GBankAddr)=1);
(*(volatile short *)(GBankAddr)=0);
(*(volatile short *)(GBankAddr)=1);
(*(volatile short *)(GBankAddr)=0);
(*(volatile short *)(GBankAddr)=1);
}
}

Test.o:    file format elf32-littlearm

Disassembly of section .text:

00000f40 <TestSysCallSpeed>:
   f40: e59f3040 ldr r3, [pc, #64] ; f88 <TestSysCallSpeed+0x48>
   f44: e59f2040 ldr r2, [pc, #64] ; f8c <TestSysCallSpeed+0x4c>
   f48: e5933000 ldr r3, [r3]
   f4c: e3a01000 mov r1, #0 ; 0x0
   f50: e1c310b0 strh r1, [r3]
   f54: e3a01001 mov r1, #1 ; 0x1
   f58: e1c310b0 strh r1, [r3]
   f5c: e3a01000 mov r1, #0 ; 0x0
   f60: e1c310b0 strh r1, [r3]
   f64: e3a01001 mov r1, #1 ; 0x1
   f68: e1c310b0 strh r1, [r3]
   f6c: e3a01000 mov r1, #0 ; 0x0
   f70: e1c310b0 strh r1, [r3]
   f74: e2522001 subs r2, r2, #1 ; 0x1
   f78: e3a01001 mov r1, #1 ; 0x1
   f7c: e1c310b0 strh r1, [r3]
   f80: 1a0003d1 bne 1ecc <TestWinDr+0x16c>
   f84: e1a0f00e mov pc, lr
   f88: 00000000 andeq r0, r0, r0
   f8c: 00895440 addeq r5, r9, r0, asr #8

-------------------------------------------------------------------------------

void TBank(void)
{

(*(volatile short *)(GBankAddr)=0);
(*(volatile short *)(GBankAddr)=1);
(*(volatile short *)(GBankAddr)=0);
(*(volatile short *)(GBankAddr)=1);
(*(volatile short *)(GBankAddr)=0);
(*(volatile short *)(GBankAddr)=1);

}

t.o:    file format elf32-littlearm

Disassembly of section .text:

00000000 <Tbank>:
0: e59f1038 ldr r1, [pc, #56] ; 40 <.text+0x40>
4: e3a00001 mov r0, #1 ; 0x1
8: e3a0c000 mov ip, #0 ; 0x0
c: e5913000 ldr r3, [r1]
  10: e1c300b0 strh r0, [r3]
  14: e5912000 ldr r2, [r1]
  18: e1c2c0b0 strh ip, [r2]
  1c: e5913000 ldr r3, [r1]
  20: e1c300b0 strh r0, [r3]
  24: e5912000 ldr r2, [r1]
  28: e1c2c0b0 strh ip, [r2]
  2c: e5913000 ldr r3, [r1]
  30: e1c300b0 strh r0, [r3]
  34: e5912000 ldr r2, [r1]
  38: e1c2c0b0 strh ip, [r2]
  3c: e12fff1e bx lr
  40: 00000000 .word 0x00000000

第一个是应用程序,第二个是驱动程序.没有太大区别.
由于CPU的主频是667M,8级流水线的arm11,所以像for(i=0;i<50;i++)这样的指令本身就执行的很快.

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snail_314

家境小康

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13楼 [报告]

发表于 2010-10-25 14:28 |只看该作者

说实话我很怀疑我的解释哈.

恩,从反汇编来看,第一个程序的r2就是i.
f8c: 00895440, f8c放的就是9000000.i的访问没涉及到内存

不过从你的代码来看,我有个疑问,你的driver中没有循环哈.而你的应用是在循环中完成的.那么不排除调度延迟.你示波器对第一个应用输出的波形整齐吗?

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zhangxujun1981

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14楼 [报告]

发表于 2010-10-25 14:33 |只看该作者

本帖最后由 zhangxujun1981 于 2010-10-25 14:40 编辑

是这样的,驱动程序没有使用循环,而是在用户程序中这样调用的:
for(i=0;i<80000000;i++){
ioctl(Gfd_misc,45,0);
}
驱动程序:
case 45:
Tbank();
break;

波形很整齐,不是随机的.应该不是调度引起的,因为只有几百ns,而且每次都是这样.我的图形只是其中的一个周期

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zhangxujun1981

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15楼 [报告]

发表于 2010-10-25 14:36 |只看该作者

struct file_operations skeleton_fops = {
.owner = THIS_MODULE,
.llseek = NULL,
.read = skeleton_read,
.write = skeleton_write,
.readdir = NULL,
.poll = NULL,
.ioctl = skeleton_ioctl,
.mmap = skeleton_mmap,
.open = skeleton_open,
.flush = NULL,
.release = skeleton_release,
.fsync = NULL,
.fasync = NULL,
.lock = NULL,
};

请问,我要在驱动程序里添加mmap函数,并且实现你说的映射吗?